Let factorials be added to their corresponding triangular numbers resulting to a sequence of numbers {2, 5, 12, 34, 135, 741, ...}, which I call factoriangular numbers in my recent papers. Let T(n) be the nth triangular number. Notice that 741 = 6! + T(6) = T(38). Are there other numbers that are both factoriangular and triangular?
To look for examples it is useful to know that the largest power of 2 dividing n! is 2^{n-1}. Apply this to the equation n! + k(k-1)/2 = m! to deduce that 2^n + k(k-1) is congruent to 0 modulo 2^m. For large n this greatly constrains the possible k.
To Stephen: not really. The power of 2 in n! is [n/2] + [n/(2^2)] + [n/(2^3)] +.... . For n = 6, this power is 3 + 1 = 4 and not 6-1 = 5. Indeed 720 = 16 x 45.
On the question: we know that all numbers which are not powers of 2 have different representations as difference of non-consecutive triangular numbers:
http://www.fq.math.ca/Scanned/39-3/nyblom.pdf
Because of this, I think that a proof for the finitude of solutions to our questions could be very hard. On the other hand, it would be really remarkable and surprizing if one had infinitely many solutions...
@Mihai: Thanks. I was way too quick. There are two formulas for the power of p in n!. The first is \sum_{k\ge1} [n/p^k] and the second is (n-s_p(n))/(p-1) where s_p(n) is the sum of the digits of the base-p expansion of n. Thus the power of 2 in n! is n-1 if and only if n is a power of 2. (The 2-part of 6! is 2^4, and 6-s_2(6)=4 which agrees with your calculation.)
If we can find positive integer solutions to 8n! + (2n+1)2 = m2, then n! + T(n) = T([m-1]/2). For example (n, m) = (6, 77) is a solution to the said equation, hence 6! + T(6) = T (38).
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